题解 | #统计每个学校各难度的用户平均刷题数#

select
    a.university,
    c.difficult_level,
    round(
        (count(b.question_id)) / (count(distinct b.device_id)),4) as avg_answer_cnt
from
    user_profile as a,
    question_practice_detail as b,
    question_detail as c
where
    a.device_id = b.device_id
    and b.question_id = c.question_id
group by
    a.university,
    c.difficult_level

having用于对分组的条件筛选

where用于对列的条件筛选

group by 放在where 之后，放在having之前

计算平均每个学校各难度的用户平均刷题数 公式要注意